feat: update solutions to lc problems: No.1170,1183

Author: yanglbme

solution/1100-1199/1170.Compare Strings by Frequency of the Smallest Character/README.md

@@ -196,18 +196,9 @@ function numSmallerByFrequency(queries: string[], words: string[]): number[] {
         for (const c of s) {
             cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
         }
-        for (const x of cnt) {
-            if (x) {
-                return x;
-            }
-        }
-        return 0;
+        return cnt.find(x => x > 0);
     };
-    const nums: number[] = [];
-    for (const w of words) {
-        nums.push(f(w));
-    }
-    nums.sort((a, b) => a - b);
+    const nums = words.map(f).sort((a, b) => a - b);
     const ans: number[] = [];
     for (const q of queries) {
         const x = f(q);

solution/1100-1199/1170.Compare Strings by Frequency of the Smallest Character/README_EN.md

@@ -174,18 +174,9 @@ function numSmallerByFrequency(queries: string[], words: string[]): number[] {
         for (const c of s) {
             cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
         }
-        for (const x of cnt) {
-            if (x) {
-                return x;
-            }
-        }
-        return 0;
+        return cnt.find(x => x > 0);
     };
-    const nums: number[] = [];
-    for (const w of words) {
-        nums.push(f(w));
-    }
-    nums.sort((a, b) => a - b);
+    const nums = words.map(f).sort((a, b) => a - b);
     const ans: number[] = [];
     for (const q of queries) {
         const x = f(q);

solution/1100-1199/1170.Compare Strings by Frequency of the Smallest Character/Solution.ts

@@ -4,18 +4,9 @@ function numSmallerByFrequency(queries: string[], words: string[]): number[] {
         for (const c of s) {
             cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
         }
-        for (const x of cnt) {
-            if (x) {
-                return x;
-            }
-        }
-        return 0;
+        return cnt.find(x => x > 0);
     };
-    const nums: number[] = [];
-    for (const w of words) {
-        nums.push(f(w));
-    }
-    nums.sort((a, b) => a - b);
+    const nums = words.map(f).sort((a, b) => a - b);
     const ans: number[] = [];
     for (const q of queries) {
         const x = f(q);

solution/1100-1199/1175.Prime Arrangements/README.md

@@ -51,7 +51,7 @@
 
 我们在 $[2,n]$ 范围内顺序遍历每个数 $i$，如果这个数为质数（$primes[i]==true$），质数个数增 1，然后将其所有的倍数 $j$ 都标记为合数（除了该质数本身），即 $primes[j]=false$，这样在运行结束的时候我们即能知道质数的个数。
 
-时间复杂度 $O(nloglogn)$。
+时间复杂度 $O(n \times \log \log n)$。
 
 <!-- tabs:start -->
 

solution/1100-1199/1183.Maximum Number of Ones/README.md

@@ -169,11 +169,7 @@ var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
         }
     }
     cnt.sort((a, b) => b - a);
-    let ans = 0;
-    for (let i = 0; i < maxOnes; ++i) {
-        ans += cnt[i];
-    }
-    return ans;
+    return cnt.slice(0, maxOnes).reduce((a, b) => a + b, 0);
 };
 ```
 

solution/1100-1199/1183.Maximum Number of Ones/README_EN.md

@@ -151,11 +151,7 @@ var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
         }
     }
     cnt.sort((a, b) => b - a);
-    let ans = 0;
-    for (let i = 0; i < maxOnes; ++i) {
-        ans += cnt[i];
-    }
-    return ans;
+    return cnt.slice(0, maxOnes).reduce((a, b) => a + b, 0);
 };
 ```
 

solution/1100-1199/1183.Maximum Number of Ones/Solution.js

@@ -15,9 +15,5 @@ var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
         }
     }
     cnt.sort((a, b) => b - a);
-    let ans = 0;
-    for (let i = 0; i < maxOnes; ++i) {
-        ans += cnt[i];
-    }
-    return ans;
+    return cnt.slice(0, maxOnes).reduce((a, b) => a + b, 0);
 };